Відео

Good Explanation

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Giving TLE

you used to have a different theme, why you changed it ? and how did u got it in the first place?

Great Explanation..Thanks

very messy writing and very messy explanation 😩

Very nice explanation👍

Pyhon code: class Solution: def countPairs(self,arr,brr): def findGreaterElements(key): low, high = 0, len(brr)-1 while low <= high: mid = (low+high)//2 if brr[mid] > key: high = mid-1 else: low = mid+1 return len(brr) - high - 1 # we need to no of elements greater # than x in brr effeciently. brr.sort() cnt = 0 one, two, threeFour = brr.count(1), brr.count(2), brr.count(3)+brr.count(4) for x in arr: # with x==1, no pair can be formed, ever. if x == 1: continue # exception: for x=2 and y=3 or 4, pairs cannot be formed if x == 2: cnt -= threeFour # exception: for x=3, y=2 can form a pair if x == 3: cnt += two # for any x and y=1, pairs can be formed cnt += one # every number greater than x can form a pair cnt += findGreaterElements(x) return cnt

How do we come up with these exceptions while in an interview?

Apki consistency 🔥🔥

bhai do you have any update about 30dayspotd challenge?

set is better suited for this, is there any performance impact on using maps instead of set ? I guess no since SC would be same

Hello Sir, we can do the same problem in O(N), using Quickselect Algorithm, Thanks for the great explanation.

I have one doubt why we are traversing array index from i=1 instead of i=0 in the last loop, since array indexing always start from 0..please clear..

Arrays.sort(arr) Return k-1

k<=0 , is the better condition, it's fine here as gfg has given unique elements..

Hello bhaiyya, I've solved this problem using the max-heap. int kthSmallest(vector<int> &arr, int k) { // code here priority_queue<int> pq; for(int i=0;i<arr.size();i++){ pq.push(arr[i]); if(pq.size()>k) pq.pop(); } return pq.top(); } Its T.C = O(n) and S.C = O(1) and, Thank you bhaiiyaa..

Thanks for Guiding me through your video , I always watch your video even i solve question myself . As, Always there is a new thing to learn from your video

nycc .. just a suggestion try to explain gfg/ leetcode contest solutions . ur views will also go up . THANKYOU

best approach i have seen! thanks for this

Thank you bhaiyya

NIT patna ❤

Check ex: n=7, x=2,y=4,z=5 Why can't we cut 7 3 times with x?? Why the ans here is 2 , by using x and z ?? Wrong problem

nice explanation 👍

If me break kyu Kara..?

Bhai ek int variable me value store karane se solve kyu nhi ho rha..?

Khudse solve kiya, and Thank you bhaiiya

Best Explanation in the entire youtube....Thanks bhaiya

double ans = Math.sqrt((double)num); return (long)ans; I liked your approach as well. Helpful for interviews when they ask to solve without inbuilt methods.

ajka potd hata diya kya?

Well Explained

public boolean isValid(String str) { String[] arrOfStr = str.split("\\."); if(arrOfStr.length != 4){ return false; } for(int i=0;i<arrOfStr.length;i++){ if(arrOfStr[i].equals("")){ return false; } if((arrOfStr[i].length() > 1) && (arrOfStr[i].charAt(0)) == '0' ){ return false; } int x=Integer.parseInt(arrOfStr[i]); if(x <= 255 ){ continue; }else{ return false; } } return true; what is problem with this

Sir What a Beautiful Explanation

if (str[index + 1] == '0' && i - index - 1 > 1) return false; if (str[index + 1] == '0' && str.length() - index - 1 > 1) return false; for checking leading zeros at start and end

i solved on my own, but they changed test cases, did few modifications now , working fine idk why they do this... btw we can use regex to solve these type of problem also.... I will try that later... and ye technically TC bhi O(1) hua an since str length is 15 max....

well and informative explain vaiya💖

Solved on my own, Thank you bhaiyya

First time I am facing this problem I submitted a wrong code which passed all tc After watching ur video I realized that I did a small wrong for which code should not run And surprisingly now i am trying to submit the old code which was accepted it is showing Rte 😅

Can we solve it using recurrsion.. ? if yes . what would be its T.C ?? Thank you bhaiyya

Aaj wala khud se bana lia bhai...Nice explaination!!! class Solution { //Function to return a list containing the bottom view of the given tree. public ArrayList <Integer> bottomView(Node root){ // Code here SortedMap<Integer, Integer> map = new TreeMap<>(); Queue<Pair> q = new LinkedList<>(); q.add(new Pair(root, 0)); while(!q.isEmpty()){ Pair p = q.poll(); Node currNode = p.node; int currLevel = p.level; if(currNode.left != null) q.add(new Pair(currNode.left, currLevel-1)); if(currNode.right != null) q.add(new Pair(currNode.right, currLevel+1)); map.put(currLevel, currNode.data); } ArrayList<Integer> ans = new ArrayList<>(); for(Map.Entry<Integer, Integer> entry:map.entrySet()){ ans.add(entry.getValue()); } return ans; } } class Pair{ Node node; int level; Pair(Node node, int level){ this.node = node; this.level = level; } }

Understood, Thank you bhaiiya

Today I solved this problem myself, thank u bhaiyya

Really Informative Bhaiya !!!!!

Easily understood, thank you bhaiyya

very neat and clean explanation bhaiya